A Note on the Normal Index and the c-Section of Maximal Subgroups of a Finite Group

The remaining notation and terminology in this paper are standard, as in Huppert [1]. In 1959, Deskins [2] introduced the concept of normal index. For a maximal subgroup M of a group G, the order of a chief factor H/K of G, where H is minimal in the set of normal supplements of M in G, is known as the normal index of M of G, denoted by η(G : M). If H/K is such a chief factor, then G = MH, K ≤ M, and |G : M| = |H/K : (H/K) ∩ (M/K)|, so |H/K| = |(H ∩ M)/K‖G : M|. The intersection (M ∩ H)/K is called a c-section of M. Li and Wang in [3] proved that every maximal subgroup M of G has a unique c-section up to isomorphism. Let Sec(M) denote a group which is isomorphic to a c-section ofM.Then η(G : M) = |Sec(M)| ⋅ |G : M|. Deskins [2] showed that G is solvable if and only if η(G : M) = |G : M| for every maximal subgroup M of G. The investigations on the normal index have been developed by many scholars; see [3–7]. But the earlier results concern the cases where p is either the largest prime dividing |G| or an odd prime. In 2010, Zhang and Li analyzed the case when p = 2 and obtained some interesting results. In particular we note the following theorems.


Introduction
In this paper, all groups considered are finite.Let () denote the set of prime divisors of ||, and for  ∈ () let Syl  () denote the set of Sylow -subgroups of .Write  ⋖  to indicate that  is a maximal subgroup of .For convenience, we cite the following relative definitions.For a fixed prime  ∈ (), (5) F  () = F  () ∩ F  ().
The remaining notation and terminology in this paper are standard, as in Huppert [1].
In 1959, Deskins [2] introduced the concept of normal index.For a maximal subgroup  of a group , the order of a chief factor / of , where  is minimal in the set of normal supplements of  in , is known as the normal index of  of , denoted by ( : ).If / is such a chief factor, then  = ,  ≤ , and | : | = |/ : (/) ∩ (/)|, so |/| = |( ∩ )/‖ : |.
The intersection ( ∩ )/ is called a -section of .Li and Wang in [3] proved that every maximal subgroup  of  has a unique -section up to isomorphism.Let Sec() denote a group which is isomorphic to a -section of .Then ( : ) = |Sec()| ⋅ | : |.Deskins [2] showed that  is solvable if and only if ( : ) = | : | for every maximal subgroup  of .The investigations on the normal index have been developed by many scholars; see [3][4][5][6][7].But the earlier results concern the cases where  is either the largest prime dividing || or an odd prime.In 2010, Zhang and Li analyzed the case when  = 2 and obtained some interesting results.In particular we note the following theorems.We observe that Theorems 1 and 2 still hold by replacing 2 with another prime .For example, let  =  4 and let  ∈ F 3 ().Since the order of  is 6 or 12, ( : ) 3 = 1.So  satisfies the hypotheses of Theorem 1.But  is 3-solvable.It is natural to ask that the theorems above hold or not for any prime .In part 3, we give positive answer and relative results.

Preliminary Results
Lemma 3 (see [8,Lemma 2.2]).Let  be a group,  a normal subgroup of , and  ∈ ().Let  be a maximal subgroup of  and  ≤ .
In view of Theorem 7 it is natural to ask if a group  is -solvable when |Sec()|  =   or 1, for  ∈ F  (), where  is a prime divisor of ||.The answer of the question is negative.For example, set  = (2, 7) and  = 3; every maximal subgroup  satisfies that |Sec()| 3 = 3, but  is not 3-solvable.For -solvable, the condition that Sec() is an abelian -group is crucial.
It is proved in [6,Theorem 7] that a group  is supersolvable if and only if, for each maximal subgroup  of , ( : )  = | : |  = 1 or .It is natural to ask if a group  is -supersolvable when ( : )  = 1 or  for any maximal subgroup  of .The answer of the question is negative.For example, set  = (2, 7) and  = 3; every maximal subgroup  satisfies that ( : ) 3 = 3, but  is not 3-supersolvable.But assuming that  = 2, the result holds or not.For the question, we give the positive answer.Next, we prove the result.Theorem 8.  is 2-supersolvable if and only if, for any maximal subgroup  of , ( : ) 2 = 1 or 2.
⇐: Conversely, assume the result is not true and let  be a counterexample of minimal order.Now, we assert  is not simple.If not, then ( : ) 2 = || 2 = 1 or 2. For || 2 = 1, it is clear that  is 2-supersolvable, a contradiction.Assume that || 2 = 2. Then  is a cyclic group of order 2, and so  is 2-supersolvable, a contradiction.This contradiction shows  is not simple.Let  be the minimal normal subgroup of .By Lemma 3, / satisfies the hypotheses of the theorem.The minimal choice of  implies that / is 2-supersolvable.If  is contained in each maximal subgroup  of , then  ⊆ Φ(), and consequently, /Φ() is 2-supersolvable, and